1 lec 19: entropy changes, relative pressures and volumes, work

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Lec 19: Entropy changes, relative pressures and volumes, work

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• For next time:– Prepare for Midterm 2 on Thursday, November

6th

• Outline:– Isentropic processes for ideal gases– Internal reversible work– Entropy balance equations

• Important points:– Try to identify the “governing equations” and

not get bogged down in all the special cases– Understand how to use the Tds relationships– Don’t forget to apply the 2nd Law when

working problems with entropy or internal reversible processes

3

Isentropic processes of ideal gases with constant specific heats

• Before we had the situation where the specific heat could be considered constant,

1

2p

T

T

p T

Tlnc

T

dTc

2

1

1

2

1

2p12 p

plnR

T

Tlncss

1

2

1

2p p

plnR

T

Tlnc

If this is zero, so that s2 = s1

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Isentropic processes of ideal gases with constant specific

heats: approximation

pc

R

1

2

1

2

p

pln

T

Tln

pc

R

1

2

1

2

p

p

T

T

Then

And

k

1k

c

R

p

The value in the exponent is:

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k

1k

1

2

1

2

p

p

T

T

So

This is only applicable for an isentropic process and cp, cv and k constant.

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1

2

1

2v12 v

vlnR

T

Tlnc0ss

1k

2

1

1

2

v

v

T

T

From the other entropy change equation for an ideal gas with constant specific heats

Simplifying, we get:

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k

2

1

1

2

v

v

p

p

k11

k22 vpvp

We can combine the two T2/T1 equations to get:

So the polytropic process with n=k is an isentropic process.

Rearrange:

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TEAMPLAYTEAMPLAY

Assuming constant specific heats, find the final pressure of air which undergoes an isentropic process from P1 = 14.7 psia, 60°F to a final temperature of 620°F.

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Isentropic processes

0p

plnR)T(s)T(sss

1

21

o2

o12

For ideal gases with variable specific heats, the calculations are straightforward:

Because it is isentropic, s2 – s1 = 0.

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1

21

o2

o

p

pRln)(Ts)(Ts

If T2 is needed and T1, p1, and p2 are known,

and one may need to interpolate to find T2.

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]R/)T(sexp[

]R/)T(sexp[

R

)T(s)T(sexp

p

p

1o

2o

1o

2o

1

2

If p1, T1, and T2 are known and p2 is needed,

Notice from this that p2 and p1 can be given as functions of T, for isentropic processes.

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(T)/R]exp[s(T)p or

So, the relative “pressure” is defined and tabulated:

Note: pr is not a pressure in spite of its name--it has no units. The right hand side is also only dependent on temperature.

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1r

2r

s1

2

p

p

p

p

We can use the pressure ratios directly for determining pressure changes in isentropic processes.

NOTE: Pr only works for isentropic processes!!! Don’t try to use it for any other process.

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TEAMPLAYTEAMPLAY

Use relative pressures in your tables in your books to find the final pressure of air which undergoes an isentropic process from P1 = 14.7 psia, 60°F to 620°F.

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1r

2r

s1

2

v

v

v

v

There is a similar relationship for volumes:

As with Pr, this only applies to isentropic relationships and the right hand side only depends on temperature.

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Work in an internally reversible flow system

• Earlier we had

• This was true for a quasiequilibrium process.

pdvw

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Work for an internally reversible flow system.

• Quasiequilibrium process--one for which departures from equilibrium are infinitesimally small.

• All states through which a system passes in a quasiequilibrium process may be considered to be themselves equilibrium states.

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Work for an internally reversible flow system

• A reversible process must proceed through a series of equilibrium states.

• Otherwise, there would be a tendency for the system to change spontaneously, which is irreversible.

• Therefore, a quasiequilibrium process is an (internally) reversible process

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• Consider an internally reversible steady flow system:

• Second law

2

1

revint Tdsq

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• The first law (not limited to internally reversible processes at this point) says

)zz(g

22hhwq 12

21

22

12 VV

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• If we do limit it to internally reversible processes, which we will emphasize with subscripts,

• Then the heat transfer term can be replaced

)zz(g

22hhwq 12

21

22

12revintrevint VV

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Work for an internally reversible flow process

• Combining the laws yields

• Rearranging

)zz(g

22hhwTds 12

21

22

12revint

2

1

VV

)zz(g

22hhTdsw 21

22

21

12

2

1

revint

VV

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• Now, use a Tds relation:• Tds = dh – vdp

2

1

2

1

2

1

vdpdhTds

2

1

12

2

1

vdp)hh(Tds

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)zz(g22

vdpw 21

22

21

2

1

revint VV

•An expression for internally reversible work in a steady flow process.

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• In the absence of KE and PE effects,

• On a P-v diagram,


2

1

revint vdpwP

2

1

v

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Work

• For open systems,

• For closed systems

2

1

revint vdpw

2

1

revint pdvw

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Compressor work

• By using the relationship Pvn = constant (or Pvk = constant) and solving for

v = , the expression

or as the book uses can be integrated to get the expressions on p. 310

n

1

P/c 2

1

revint vdpw

2

1

revint vdpw

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Entropy change for a closed system

genST

QSS

2

1

12

Entropy change of system as it goes from 1 to 2; can be + or – depending on the other two terms.

Entropy trans-fer to the sys-tem via heat transfer; can be + or –, dep-ending on the sign for Q.

Entropy prod-uction term: >0 when internal irrerversibilities are present; =0 when no int irr are present; <0 never.

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Entropy change of an internally reversible process; heat transfer

T

s

P=c

V=c

1

2The area under either the red or blue curves would represent the heat transfer for either process:

2

1

12 Tdsq

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TEAMPLAYTEAMPLAY

What does a reversible, adiabatic process look like on a T-s diagram??

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Entropy production and transfer

We can use entropy (the second law) to find the minimum work (associated with a reversible process).

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Example Problem

Refrigerant 134a is compressed in a piston-cylinder assembly from saturated vapor at 10°F to a final pressure of 120 psia. Determine the minimum theoretical work input required per unit mass of refrigerant, in Btu/lb.

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Look at possibilities on a Ts diagram...

T, F

S

P = 26.65 psia

P = 120 psia

10 1

2?

2?

2?

Starting at 1, where is point 2? What can we know?

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Adiabatic Compression of R-134a

PEKEuuwq 12 The first law gives us

We know the process is adiabatic, so q = 0.

We ignore KE and PE.

Then w = u1 – u2

u1 = 94.68 Btu/lb from Table A-11E.

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21 uuw

We need to find u2 for minimum work on this compressor.A reversible system will have minimum work done on it. We showed this in ch. 6 for cycles. It is shown again for open systems on pp. 308-309. That analysis can easily be extended to closed systems by replacing h with u.

Minimum work corresponds to the smallest allowable value for u2, which we can determine from the second law.

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gen12 sT

qss

• q = 0 because it is adiabatic.

• sgen = 0 because the minimum work occurs for a reversible process (frictionless).

• So s2 = s1

• For irreversibilities, sgen > 0, which implies that s2 > s1

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T, F

S

P = 26.65 psia

P = 120 psia

101

2s

2a

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TEAMPLAYTEAMPLAY

The reversible situation is s2 = s1 = 0.2214 Btu/(lbm-R)

For a simple, compressible system, you now know enough to find u2s and w.

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Entropy rate balance for control volumes

genj j

j ST

Q

dt

dS

genb

ST

QSS 12

For a closed system, we had (as one of our forms of the second law)

For a closed system, a rate form can be written:

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geneexite

eiinleti

ij j

jcv SsmsmT

Q

dt

dS

Now, if entropy is transferred or convected, into and out of the control volume, the equation becomes, for an open system

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Rate of change of entropy within the control volume (CV).

Rate of entropy transfer into the CV as a result of heat transfer

Rate of entropy transfer into the CV with the mass flow.

Rate of entropy transfer out of the CV with the mass flow.

Rate of entropy produc-tion due to irre-versibili-ties in the CV

geneexite

eiinleti

ij j

jcv SsmsmT

Q

dt

dS

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genei SssmT

Q

)(0

Assume steady flow, one heat transfer term , and one exit and one inlet:

geneexite

eiinleti

ij j

jcv SsmsmT

Q

dt

dS

This can be simplified:

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0 ( )i e gen

qs s s

T

mNow, if one divides through by

we get

2 1e i gen

qs s s s s

T

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•This equation is almost identical to the one for closed systems. In that case, the whole system goes between two states, 1 and 2.

•For an open system, the change is from inlet to exit, and it is for the change of condition of the mass in the system as it moves from state 1 at the inlet to state 2 at the exit.

2 1 gen

qs s s

T

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For sgen 0 (irreversibilities present):

If the process is also adiabatic, then

s2 s1

What can we conclude? An adiabatic, irreversible process will always move to the right on a Ts diagram.

2 1

qs s

T

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Look at adiabatic compression processes on a Ts diagram

ForbiddenStates

T

s

P1

P2

1

2s 2a

P2 > P1

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Look at adiabatic expansion processes on a Ts diagram

ForbiddenStates

T

s

P1

P2

1

2s 2a

P1 > P2

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Isentropic Processes(constant entropy)

For real substances, such as water, R-134a, etc., use tabular entropy values or EES.

An isentropic compressor has R-134a saturated vapor entering at 10°F and leaving the compressor at 120 psia. What does the process look like on a Ts diagram?

49


T, F

S

P = 26.65 psia

P = 120 psia

101

2s

Pt 2 is at s2 = s1 and p = 120 lbf/in2.

1 lec 19: entropy changes, relative pressures and volumes, work

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